AoPS Academy
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, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
, prove that 
where 
be positive real numbers such that 
Prove that:
be the set of integers. Determine all functions 
such that, for all integers 
and 
, 
Proposed by Liam Baker, South Africa
such that has a multiple which is alternating.
be an acute-angled triangle with 
. Let 
be the orthocenter of triangle 
, and let 
be the midpoint of the side 
. Let 
be a point on the side 
and 
a point on the side 
such that 
and the points , , are on the same line. Prove that the line 
is perpendicular to the common chord of the circumscribed circles of triangle and triangle 
.
, 
, and 
are fixed on a line in this order. Let 
be a circle passing through and whose center does not lie on the line . Denote by 
the intersection of the tangents to at and . Suppose meets the segment 
at 
. Prove that the intersection of the bisector of 
and the line does not depend on the choice of .
and 
be two points inside triangle 
such that![\[ \angle MAB = \angle NAC\quad \mbox{and}\quad \angle MBA = \angle NBC.
\]](http://latex.artofproblemsolving.com/2/0/4/20410178c70693881f348c67454f01057a3fc40c.png)
![\[ \frac {AM \cdot AN}{AB \cdot AC} + \frac {BM \cdot BN}{BA \cdot BC} + \frac {CM \cdot CN}{CA \cdot CB} = 1.
\]](http://latex.artofproblemsolving.com/b/b/6/bb64f80eccb4634b9c9e15cade2f7c0b87b60923.png)
the set of all positive real numbers.
which have the property:![\[f(x)f(y)=2f(x+yf(x))\]](http://latex.artofproblemsolving.com/4/f/f/4ffbde35ff082585a188c1a4368d39338d1b07a4.png)
and 
.
questions, each with 
answer choices. A participant who answers all questions correctly will receive a gift. To ensure that at least one member of my family answers all questions correctly, how many family members need to take the quiz?
, each of our twin children also begins to have twins every year. If this pattern continues, how many years will it take for my family to grow large enough to have the required number of members to guarantee winning the quiz gift?
, the following inequality holds:
When does equality hold?
with 
, the foot of altitudes from 
to the sides 
are 
, respectively. is the orthocenter. is the midpoint of segment . Lines 
and 
intersect at 
. Let the tangents drawn to circumcircle 
from and intersect at 
. Prove that 
are colinear
be an integer and let 
be positive integers. In a group of 
people, some games of chess are played. Two people can play each other at most once. Prove that it is possible for the following two conditions to hold at the same time:
.
with 
, there is someone who has played exactly 
games of chess.
, 
, and![\[
a_{2k+1} = 2 + 2a_k, \quad a_{2k+2} = 2 + a_k + a_{k+1},
\]](http://latex.artofproblemsolving.com/4/e/c/4ec892df19b4ec97af9bb1e5d1954e6c26be3172.png)
for all integers 
. Determine all positive integers 
such that![\[
\frac{a_n}{n}
\]](http://latex.artofproblemsolving.com/3/d/1/3d1d012e7c41e1e09abd14626a79b1fb1cbab34d.png)
is an integer.
is odd. Now calculate some values for 
you can see that for 
type numbers 
is an integer. Then just simple induction proves the result.
, , and![\[
a_{2k+1} = 2 + a_k, \quad a_{2k+2} = 2 + a_k + a_{k+1},
\]](http://latex.artofproblemsolving.com/4/e/d/4edb78766605f6eb84f2822706527ef16b0e11ef.png)
for all integers . Determine all positive integers such thatis an integer.
, , andfor all integers . Determine all positive integers such thatis an integer.
for all , then only possible integer solution for 
is or 
. is obvious , due to the solution present by others, also values of new sequence you give will be smaller than or equal to the original sequence.
Hatt's off to the proposer of Problem 6 in INMO 2025. The only problem I fail to solve in the whole paper. I was near but didn't observe certain things. I only solved the case when 
but couldn't go further and finally saw the solution in aops.
for all 
. The following is the key claim.,![\[n<a_n \le 2n\]](http://latex.artofproblemsolving.com/d/0/b/d0b8b50120725a3f797b06c6cd527bdb785a366a.png)
trivially, 
and 
so the base cases are clear. Now, say the claim holds for all 
for some 
. Then,![\[2k+1 < 2+2k<2+2a_k=a_{2k+1} = 2+2a_k \le 2+2(2k)=2(2k+1)\]](http://latex.artofproblemsolving.com/1/6/6/16634888e5890ded272d675dba07d59bc8178b0a.png)
with equality on the upper bound if and only if 
. Similarly,![\[2k+2 < 2+k + (k+1) < 2 + a_k + a_{k+1} = a_{2k+2} = 2 + a_k + a_{k+1} \le 2+2k + 2(k+1)= 2(2k+2)\]](http://latex.artofproblemsolving.com/b/1/b/b1bd3d3247e313fbcd9ff09858b8d44d9dd32871.png)
with equality on the upper bound if and only if and 
. is an integer, the claim implies that we require 
. Now, consider the minimal even 
such that 
. As a result of the equality condition noted above, we require 
and 
(since both of these are indeed valid terms). But these are consecutive terms, so one of them must have an even index. But since![\[\frac{n-2}{2} < \frac{n-2}{2}+1 < n\]](http://latex.artofproblemsolving.com/3/a/9/3a9ca2ae73e4f103b2f0fad5fcb3a7acc8fc3364.png)
for all , this implies that there exists a smaller even indexed term 
which is a contradiction. Thus, for all even , is not an integer., 
if and only if 
as a result of the equality condition. We again resort to induction. Say the only integers 
for some 
such that are those of the form for 
. Then, for 
, if and only if . But,![\[2^{k-1}-1=\frac{2^k -2}{2}<\frac{n-1}{2} \le \frac{2^{k+1}-2}{2}=2^k -1\]](http://latex.artofproblemsolving.com/e/0/a/e0a91b92240920dbd2845789c8b114a526308c3f.png)
and the only such term in this range is 
by assumption. Thus, the only such term in the range is 
which completes the induction.
for some positive integer .
, , andfor all integers . Determine all positive integers such thatis an integer. for all , then only possible integer solution for is or . is obvious , due to the solution present by others, also values of new sequence you give will be smaller than or equal to the original sequence.
for some bigger k and the k is not that much big and 
so the only thing that I have to care is that 
and yeah that's the place where I stuck how to prove if that is true or false. If it is false then how can I get to the contradiction
- 1
.
,
or 
.
from 
to 
and 
otherwise.
for all 
and for the other interval using IVT we can show that 
,
.
,
and thus 
which is indeed true!
,
.
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,
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of the form 
and 
.
.
.
then, 
.
then, 
.
then, 
.
then, 
.
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,
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:
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for all 
for all 
and 
is an integer in that given range in each case.
for all 
Implying, 
Thus, 
implying 
Thus by 
we have 
thus working 
for every positive integer 
for all 
Thus, 
, so, 
but 
by 
.
for all positive integer 
I checked cases upto 
.
.
, then the pattern is obvious and now induction. The only hurdle is 
mins of finding the values of the sequence till 
for all positive integers 
where 
.
by our inductive hypothesis
\hline
for some positive integer 
i
.
,
is even and other is odd, so by our inductive hypothesis,
is odd and the other is even.
. Hence our induction is complete.
is good.
.
must be good, similarly 
must also be good and so on.......
for all 
.
.
.
.
,
as desired, so our claim is proved.
work.
for all positive integers 
,
.![$[1,i]$](http://latex.artofproblemsolving.com/8/4/a/84a74ff446b0b9f0557add86de63bca29afb4f48.png)
.
,
by our hypothesis.
.
![$a_n \in (n, 2n]$](http://latex.artofproblemsolving.com/5/c/e/5ce1ea1ba1a13d3d5be44a90144a24dc75ed8b1b.png)
.
can't both be one less than powers of two. In the second equation, 
has equality, which gives the conclusion.
is odd.
be odd.
.
is odd.
.
.
.
.
and 
,